Using Gauss's law in electrostatics, obtain an expression for electric field due to a uniformly charged thin spherical shell at a point
(i) Outside the shell and
(ii) Inside the shell

Explanation

Gauss.s theorem : The total outward electric flux passing through a closed surface in air is `(1/epsilon_0)` times the total charge enclosed by it . .

Let .+Q. be the charge enclosed by a hollow conductor of radius .R..
Let .p. be a point at a distance .r. from the centre of the conductor and .ds. be a small element of area surrounding the point . A normal drawn from the point coincides with the direction of `vecE`. Hence cos `0^@=1`.
From Gauss.s theorem ,
`TOEF=phi=(1/epsilon_0)Q`...(1)
by definition f=E cos q Sds.
where, `ads=4pr^2 , cos q=cos 0^@ =1`
hence, f=E . `4pr^2`...(2)
Comparing (1) & (2)
we write `E=(1/(4piepsilon_0))Q/r^2`
For a point on the surface , r =R
`E=(1/(4piepsilon_0))(Q/R^2)`
This electric field intensity is maximum . Since electric flux depends on the charge enclosed and electric field intensity depends on the electric flux , electric field remains zero at all points inside the spherical hollow conductor.